Question

An infinitely long uniform line charge distribution of charge per unit length $\lambda$ lies parallel to the y-axis in the y-z plane at $z=\frac{\sqrt{3}}{2}a$. If the magnitude of the flux of the electric field through the rectangular surface ABCD lying in the x-y plane with its centre at the origin is $\frac{\lambda L}{n{\epsilon }_0}$ (${\epsilon }_0=$ permittivity of free space), then the value of n is :

Answer 1

Let the perpendicular from O to the line charge intersects the line charge at point Q.

$OQ=\frac{\sqrt{3}a}{2}$

Angle subtended by AB at Q is $\angle AQB=2\angle AQO$

$\tan \angle AQO=\frac{AO}{QO}=\frac{\frac{a}{2}}{\frac{\sqrt{3a}}{2}}=\frac{1}{\sqrt{3}}=\tan {30}^0$

$\therefore \angle AQB={60}^0$

Let us consider a cylindrical gaussian surface with the axis along line charge of length L and radius a(since AQ =a ).

${\varphi }_{cylinder,curved}=\frac{q_{enc}}{{ϵ}_0}=\frac{\lambda L}{{ϵ}_0}$

Flux through the rectangle is same as the flux through $\frac{1}{6}$th of the cuboid since it subtends ${60}^o$ at Q and field lines are symmetrically radially outwards.

Hence, ${\varphi }_{rectangle}=\frac{1}{6}\frac{\lambda L}{{ϵ}_0}$

$\therefore n=6$